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Let $f:\mathbb{R}\rightarrow\omega_1$ be the "usual" surjection (send $r$ to the ordinal coded by $r$ according to some fixed reasonable coding system if such an ordinal exists, and to $0$ otherwise). Consider the following statement:

$(\%)\quad$ If $G\subseteq\omega_1^\omega$ is determined (as a game on $\omega_1$) then the "preimage" $\hat{G}=\{s\in\mathbb{R}^\omega: (f(s_i))_{i\in\omega}\in G\}$ is determined (as a game on $\mathbb{R}$).

This principle follows trivially from $\mathsf{ZF+AD}_\mathbb{R}$ or $\mathsf{ZFC}$. However, the $\mathsf{ZF+AD}$ situation is unclear to me: if we let $\Sigma$ be a winning strategy for (say) player $1$ in $G$, there is no obvious way to convert it to a winning strategy for $1$ in $\hat{G}$ since that would require a canonical choice of code for each ordinal.

Question: Does $\mathsf{ZF+AD}\vdash(\%)$?

I'm happy to look instead at mild strengthenings of $\mathsf{ZF+AD}$ (e.g. add $\mathsf{V=L(\mathbb{R})}$ or $\mathsf{AD}^+$).

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Yes, $\mathrm{ZF}+\mathrm{AD}$ proves $(\%)$. The reason is that $\omega_1$ is small compared to $\Theta$, so the relevant sets of reals are well within the region in which uniformization holds.

Claim: For any $g\colon\omega_1\rightarrow\omega_1$ there is $h\colon\mathbb R\rightarrow \mathbb R$ with $f\circ h=g\circ f$.

Proof: $\omega_1$ is the first projective ordinal, so by Moschovakis' coding theorem, the set $$A=\{(x, y)\in\mathbb R^2\mid g(f(x))=f(y)\}$$ is projective and hence has a uniformization $h$. $\Box$

Similarly, for any strategy $\tau$ on $\omega_1$, there is a strategy $\sigma$ on $\mathbb R$ satisfying $f(\sigma(x_0,\dots, x_n))=\tau(f(x_0),\dots, f(x_n))$.

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